Problem: Divide the following complex numbers. $ \dfrac{-2+8i}{3+5i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3-5i}$ $ \dfrac{-2+8i}{3+5i} = \dfrac{-2+8i}{3+5i} \cdot \dfrac{{3-5i}}{{3-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-2+8i) \cdot (3-5i)} {(3+5i) \cdot (3-5i)} = \dfrac{(-2+8i) \cdot (3-5i)} {3^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-2+8i) \cdot (3-5i)} {(3)^2 - (5i)^2} = $ $ \dfrac{(-2+8i) \cdot (3-5i)} {9 + 25} = $ $ \dfrac{(-2+8i) \cdot (3-5i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2+8i}) \cdot ({3-5i})} {34} = $ $ \dfrac{{-2} \cdot {3} + {8} \cdot {3 i} + {-2} \cdot {-5 i} + {8} \cdot {-5 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-6 + 24i + 10i - 40 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-6 + 24i + 10i + 40} {34} = \dfrac{34 + 34i} {34} = 1+i $